Re: [xsl] xalan:nodeset or xsl:copy-of problem

Subject: Re: [xsl] xalan:nodeset or xsl:copy-of problem
From: Greg Faron <gfaron@xxxxxxxxxxxxxxxxxx>
Date: Tue, 23 Apr 2002 17:23:58 -0600
At 04:24 PM 4/23/2002, you wrote:
 <xsl:template name="setup_db">
    <xsl:param name="field"/>
    <xsl:param name="db_list"/>
    <xsl:param name="full_path"/>
    <xsl:variable name="nodetree">
      <xsl:for-each select="$full_path[contains($dbs_searched,@db)]">
<!-- $dbs_searched variable is passed from a different stylesheet -->
        <xsl:copy-of select="."/>
      </xsl:for-each>
    </xsl:variable>

<!-- This statement does print out the attribute value of field $field,
which means that $full_path is a valid path and does contain @db
attribute -->
    <xsl:value-of select="$full_path/@db" />

Okay, I'm re-reading your post here and the above isn't clear. Assuming that $full_path is a valid node-set, then the above selection chooses all top-level elements of $full_path that themselves have a "db" attribute. It says nothing of any child elements named $field. I think your bug is in your selection paths, not the node-set or copy-of functions.
Instead of testing with the value-of template (which concatenates all of the child nodes as well as the top-level node), use copy-of to debug your stylesheet. I think you'll find that your variable values aren't what you expect.



Greg Faron Integre Technical Publishing Co.



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